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If the breaking strength of a rod of diameter 2 cm is `2xx10^(5)N` than that for a rod of same material and diameter 1 cm will be
A. `2xx10^(5) N`
B. `1xx10^(5)N`
C. `0.5 xx 10^(5)N`
D. `0.25 xx 10^(5)N`

1 Answer

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Best answer
Correct Answer - C
`(F_(2))/(F_(1))=(r_(2)^(2))/(r_(1)^(2))=(1)/(4)`
`F_(2)=(1)/(4)xxF_(1)`
`=(1)/(4)xx2xx10^(5)=0.5xx10^(5)N`.

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